Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone please explain to me how to prove this:

$\sin2\alpha = \frac{2\tan \alpha}{1+\tan^2 \alpha}$

Also: $\alpha \neq (2k + 1)\pi, k \in \mathbb{N}$

Thanks in advance.

share|cite|improve this question
up vote 3 down vote accepted

$$\sin2\alpha=2\cos\alpha \sin\alpha=\dfrac{2\cos^2\alpha \sin\alpha}{\cos\alpha}=\dfrac{2\tan\alpha}{\sec^2\alpha}=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$$

share|cite|improve this answer

Use the fact that $\sin{2a}=2\sin{a}\cos{a}$. Now what happens if you multiply by 1?

$\mathbf{hint}: 1=\frac{\cos{a}}{\cos{a}}$...

share|cite|improve this answer

Hint: Use the fact that $\sin 2\alpha = 2\sin \alpha \cos \alpha$ and $\sin^2 \alpha + \cos^2 \alpha = 1$. Use the last hint as the denominator of a fraction.

share|cite|improve this answer

Right hand side= 2 $Tanx/(1+tan^2x)$ substitute Tanx=$sinx/cosx$ and 1+$tan^2x$= $sec^2x$

2$(sinx/cosx)$/$sec^2x$= 2$(sinx/cosx)$ * $cos^2x$ = 2sinx* cosx + sin2x (left hand side )


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.